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It is well-known that for $n \geq 2$ and a finite field $k$, the Milnor $K$-group $K_n ^M (k)$ vanishes. I don't know who proved this first, but if curious, you may look at somewhere in Srinivas's book "Algebraic K-theory" or various lecture notes in the internet.

My question is whether it is always nonzero, when $k$ is infinite. Certainly we have concrete computations in the literature where they are nonzero, and results such as Nesterenko-Suslin-Totaro, which shows Milnor K-groups are higher Chow groups of $k$ in the Milnor range. But, they don't seem to imply anything about non-vanishing of $K_n ^M (k)$ for $n \geq 2$.

It is also hard to prove that some group is nonzero even though you know the generators and relations. (In a sense, if true, one "could" possibly prove something is zero by computations, but something is "nonzero" seems to require something else.

Does anyone have some ideas about it?

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    $\begingroup$ It follows from Borel's work on the stable cohomology of artihmetic groups that $K_n^M(k)$ is always torsion for $n>1$ if $k$ is a number field. $\endgroup$
    – naf
    May 14, 2016 at 4:13
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    $\begingroup$ Also, if Milnor $K$-theory vanishes for finite fields then it clearly vanishes for any algebraic extension of a finite field (which can be infinite...). $\endgroup$
    – naf
    May 14, 2016 at 4:22
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    $\begingroup$ Are you asking for nonvanishing of all $K_n^M$, or just if there is some nonvanishing group? If $k$ is a finite field and $k(t)$ is the field of rational functions over it, then there's a Milnor exact sequence which, in particular, shows that $K^M_n(k(t))$ is trivial for $n \geq 3$. (I believe that this exact sequence is discussed at length in Bass and Tate's "The Milnor ring of a global field.") $\endgroup$
    – Tyler Lawson
    May 14, 2016 at 5:53
  • $\begingroup$ Thank you very much! I wanted to see some concrete examples where the Milnor K-groups could still be zero when $k$ is infinite. It looks I was just given in the above at least two examples where the Milnor K-groups still become zero when $k$ is algebraic over the prime subfield, and when $k$ is not algebraic over the prime subfield. $\endgroup$
    – Jinhyun Park
    May 14, 2016 at 7:06
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    $\begingroup$ For a number field $k$, $K_2(k)$ is always torsion and non-zero and has no non-zero divisible subgroups (by a theorem of Garland). If $k$ has no real embeddings, it seems likely that it follows that $K_n(k)$ is zero for $n>2$ (because $k$ has cohomological dimension 2). Moreover, $K_n(k)$ is non-zero for all $n>2$ if $k$ has a real embedding (because $K_n(k)/2$ is non-zero). $\endgroup$
    – naf
    May 16, 2016 at 7:22

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